Question

A car is traveling at 80 km/h while he sees a tractor 50m away which is traveling at 20 km/h. What should be the deceleration in order to avoid the collision.

Answer 1

Assume that the tractor and the car are moving in the same direction. If the tractor keeps moving at the same speed, the car need to decelerate at a minimum rate of approximately
`2.8\; {\rm m\cdot s^(-2)}` (
`(25 / 9)\; {\rm m\cdot s^(-2)}`) to avoid collision.

Step-by-step explanation:

Relative to the tractor, the car was initially moving at
`80\; {\rm km \cdot h^(-1)} - 20\; {\rm km \cdot h^(-1)} = 60\; {\rm km \cdot h^(-1)}`.

Apply unit conversion; ensure that the unit of velocity is
`{\rm m\cdot s^(-1)}`:

`\begin{aligned} 60\; {\rm km \cdot h^(-1)} &= 60\; {\rm km \cdot h^(-1)} * \frac{1\; {\rm m\cdot s^(-1)}}{3.6\; {\rm km \cdot h^(-1)}} \\ &\approx \; 16.7\; {\rm m\cdot s^(-1) \end{aligned}`.

In other words, the initial velocity of the car was
`u \approx 16.7\; {\rm m\cdot s^(-1)}` relative to the tractor.

Since the tractor is moving at constant velocity, the acceleration
`a` of the car relative to the tractor is the same as the acceleration of the car relative to the ground.

It is given that the initial distance between the tractor and the car was
`50\; {\rm m}`. The final velocity
`v` of the car should be no more than
`0\; {\rm m\cdot s^(-1)}`. Otherwise, the car would keep moving toward the tractor until the two vehicles collide.

Relative to the tractor, if the deceleration of the car was at the minimum safe value:

• Initial velocity of the car was
  `u \approx 16.7\; {\rm m\cdot s^(-1)}` (
  `60\; {\rm km \cdot h^(-1)}`, relative to the tractor.)
• Final velocity of the car would be
  `v = 0\; {\rm m\cdot s^(-1)}` relative to the tractor.
• Displacement of the car would be
  `x = 50\; {\rm m}` (again, relative to the tractor.)

Apply the SUVAT equation
`a = (v^(2) - u^(2)) / (2\, x)` to find the acceleration of the car:

`\begin{aligned}a &= (v^(2) - u^(2))/(2\, x) \\ &\approx \frac{(0\; {\rm m\cdot s^(-1)})^(2) - (16.7\; {\rm m\cdot s^(-2)})^(2)}{2 * 50\; {\rm m}} \\ &\approx (-2.8)\; {\rm m\cdot s^(-2) \end{aligned}`.

In other words, the deceleration of the car should be at least
`2.8\; {\rm m\cdot s^(-2)}` relative to the tractor and to the ground.